Problem: What is the extraneous solution to these equations? $\dfrac{x^2 + 12}{x + 6} = \dfrac{-9x - 6}{x + 6}$
Solution: Multiply both sides by $x + 6$ $ \dfrac{x^2 + 12}{x + 6} (x + 6) = \dfrac{-9x - 6}{x + 6} (x + 6)$ $ x^2 + 12 = -9x - 6$ Subtract $-9x - 6$ from both sides: $ x^2 + 12 - (-9x - 6) = -9x - 6 - (-9x - 6)$ $ x^2 + 12 + 9x + 6 = 0$ $ x^2 + 18 + 9x = 0$ Factor the expression: $ (x + 3)(x + 6) = 0$ Therefore $x = -3$ or $x = -6$ At $x = -6$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -6$, it is an extraneous solution.